LESSON NOTES · 01

Why the canonical distribution appears

Prerequisites: Probability and elementary differentiation

Notation: \(\beta=1/(k_B T)\) is inverse temperature, \(k_B\) is Boltzmann’s constant, and \(T\) is absolute temperature. Thus \(\beta E\) is dimensionless.

1. Count compatible bath states

A small system with energy \(E(x)\) weakly couples to a large bath in an isolated total system. Probability is proportional to compatible bath multiplicity. Expand the bath entropy around its large energy; its derivative is \(1/T\). This produces Boltzmann weights. Neglected higher terms express finite heat-capacity corrections.

\[ S_B=k_B\ln\Omega_B,\quad\ln\Omega_B(E_{tot}-E)\simeq\ln\Omega_B(E_{tot})-\beta E,\quad\pi(x)=\frac{e^{-\beta E(x)}}Z,\quad Z=\sum_xe^{-\beta E(x)}. \]

2. Configurational equilibrium

Continuous states require the appropriate phase-space integral. For separable Gaussian momenta and coordinate-only observables, integrate momenta out to leave \(\pi(R)\propto e^{-\beta U(R)}\). Degeneracy counts states and cannot be discarded.

3. Why samples can replace a huge sum

A suitable Markov chain visits regions with their equilibrium weights without evaluating \(Z\). Average the observable along the saved chain, including repeated states. These moves explore configuration space; absent a justified kinetic model, steps and sweeps are not measured seconds. Stationarity, mixing and correlated uncertainty must be checked separately.

\[ \langle A\rangle=\sum_x\pi(x)A(x),\qquad\bar A_M=\frac1M\sum_{s=1}^M A(x_s). \]

Teaching schematic drawn from the equations or algorithm steps in this lesson; not measured data.

Worked two-level probabilities

Two nondegenerate states have \(E_0=0,E_1=\varepsilon\). At \(\beta\varepsilon=\ln3\), \(Z=4/3\), excited probability is \(1/4\) and mean energy \(\varepsilon/4\). If the excited level has degeneracy \(g\), count all its states. These are analytic equilibrium probabilities, not measured samples.

\[ P_1=\frac{g e^{-\beta\varepsilon}}{1+g e^{-\beta\varepsilon}}. \]

Check yourself and limits

Is the most probable microstate necessarily the most probable macroscopic region?

Answer and reasoning

No; a region may contain many individually less-probable states.

Does random sampling make a classical model quantum?

Answer and reasoning

No; the target distribution and estimator determine the physics.

References and further reading

• Metropolis et al. (1953), canonical sampling

Derivations and numerical examples here are original teaching constructions, not copied passages or reported research data.

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